Why does oversampling make the job of an anti-alias filter easier? I don't understand the intuition on why it becomes helpful.

Isn't it economical to just keep the sampling rate at the minimal level, that is twice the highest frequency of the signal, in that way, we can save energy, and processing and computational power.


1 Answer 1


You might remember Nyquist's theorem. Given a signal which is band limited to $f_1$, we must sample it at least at $2f_1$:


So if you check you favourite Signals and Systems book (e.g. Oppenheim's), you might recall that, once sampled, we can consider the signal's discrete Spectrum (which is periodic every $2\pi$ radians, by the way).


(In the image, Hz are considered instead of radians, being a 2*pi scale factor the only difference)

Then, we would go back to continuous time by low pass filtering with $f_{cutoff}=f_S/2$. So let's look into a few cases:

  • if $f_S<f_1$, the original spectrum and the subsequent replicas would overlap ($f_1>f_S-f_1$). This would result in distortion over the original signal after the low pass filter. This is the so called aliasing.

  • if $f_S>f_1$, as in the image, there is no overlapping, and therefore no aliasing. Furthermore, we can recover the original signal by applying a low pass filter. Fortunately, we don't need the sharpest transition band in our filter.


(please excuse my lack of accuracy here, the filter would also be periodic as we still are in discrete frequency domain. Also, please take into account that 3 dB cutoff should be on $f_S/2$)

  • Last, imagine how this would be if we adjusted $f_S=f_1$. Both spectra would converge at $f_S=f_1$, so we would need an ideal (infinite slope) filter to recover the undistorted original signal. Ideal is no good, you know.

Now, let's think about those LP filters. https://i.stack.imgur.com/B7U04.png (wikipedia)

The wider the transition band is, the flatter the pass band is. On the other hand, the pass band is curlier when the transition band is tighter. Non-flat frequency response means distortion. Don't forget that the signal that we are "deleting" in the stop band would produce distortion, both in the anti-alias and the D/A filter.

So we want to keep the sampling frequency as low as possible, while not distorting the signal, either by the filter ripple or by aliasing. But, on the other hand, we want to keep the filter as cheap and accurate as possible. By allowing a wider transition band, we avoid the need of incrementing the order of the filters, i.e., the number of filter coefficients (in the D/A filter) or of electronic devices (in the anti-alias).

So we confront a cost/distortion trade-off, in which it seems preferable to increase sampling rate (e.g. CD audio is sampled at 44.1 kHz while we hear no more than 20 kHz), so we can keep reasonable distortion and avoid huge order filters.

Take into account that increasing filter's order also leads to an increment in number of operations, although it is less than the one produced by increasing sampling rate.

Filter terminology here, just in case: http://www.dspguru.com/dsp/reference/filter-terminology

  • 2
    $\begingroup$ Welcome to DSP.SE! This is a great answer! $\endgroup$
    – Phonon
    Dec 31, 2012 at 7:26

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